Chance of someone having pocket pairs in full ring - Poker Theory

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Chance of someone having pocket pairs in full ring

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08-14-2020 , 07:47 PM

Carla_

newbie

Join Date: Aug 2020 Posts: 33

I'm curious of the chance of at least 1 person having a pocket pair in texas holdem, in a full ring (9 player) table (or in more general for any number of players).
Sorry if this wrong subforum for this question, i can't think of an easy way to calculate/estimate this.

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08-15-2020 , 12:19 AM

whosnext

Carpal \'Tunnel

Join Date: Mar 2009 Posts: 6,733

There's a stickied thread in the Probability Forum which has all the information you seek.

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08-15-2020 , 04:13 AM

hyperknit

veteran

Join Date: Dec 2016 Posts: 2,692

The probability of being felt a pocket pair from 2 random cards is:
3/51
And then there’s 9 people who each get a shot at those odds.

So

(48/51)^9 = .58

So at least one guy gets a PP 42%

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08-15-2020 , 09:10 AM

Carla_

newbie

Join Date: Aug 2020 Posts: 33

Quote:

Originally Posted by hyperknit

The probability of being felt a pocket pair from 2 random cards is:
3/51
And then there’s 9 people who each get a shot at those odds.

So

(48/51)^9 = .58

So at least one guy gets a PP 42%

This is wrong, suppose one of the players doesn't have a pocket pair, then the chance of another player having one changes

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08-15-2020 , 10:07 AM

Carla_

newbie

Join Date: Aug 2020 Posts: 33

Quote:

Originally Posted by whosnext

There's a stickied thread in the Probability Forum which has all the information you seek.

Thanks, is there a way to delete this thread?

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08-15-2020 , 02:31 PM

hyperknit

veteran

Join Date: Dec 2016 Posts: 2,692

Quote:

Originally Posted by Carla_

This is wrong, suppose one of the players doesn't have a pocket pair, then the chance of another player having one changes

Does it tho?

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08-15-2020 , 07:59 PM

hyperknit

veteran

Join Date: Dec 2016 Posts: 2,692

Suppose player 1 gets delt a 5 and a 7.
Now we update the remaining deck of 50 cards to not include those.
So the way to get a pair is either:
Case A 3/49
Or
Case B 2/49

Case A occurs 44 times
Case B occurs 6 times.
So
(44/50)(3/49)+ (6/50)(2/49) = .0588

And of course 3/51 = .0588

Same thing

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08-15-2020 , 10:39 PM

whosnext

Carpal \'Tunnel

Join Date: Mar 2009 Posts: 6,733

This is probably a waste of time, but every post of yours in this thread has been wrong.

Why don't you pore over your last post and see if you can spot the error.

ETA: Consider two players and if the deck only had Aces and Kings. Then it is obvious that whether or not one player has a pair affects the prob of the other player having a pair.

-

Last edited by whosnext; 08-15-2020 at 10:56 PM. Reason: added eta

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08-16-2020 , 05:53 AM

hyperknit

veteran

Join Date: Dec 2016 Posts: 2,692

Quote:

Originally Posted by whosnext

I just read ur thread in the probability section and I liked it a lot. Ur def right about the removal effects, but also u did like 10x more math than me and we got pretty damn similar answers so I’m not sweating it.

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08-17-2020 , 09:30 PM

#10

MrDownswing

newbie

Join Date: Aug 2020 Posts: 20

What is the use of these type of stats in game? Does it really help? (Yeah, I am new

)

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08-18-2020 , 06:40 AM

#11

heehaww

Pooh-Bah

Join Date: Aug 2011 Posts: 5,081

Quote:

Originally Posted by MrDownswing

What is the use of these type of stats in game? Does it really help?

No, these stats aren't useful, partly because your opponents aren't playing blindly or randomly, which means the real probability changes the moment they make a decision.

Quote:

Originally Posted by hyperknit

(44/50)(3/49)+ (6/50)(2/49) = .0588

And of course 3/51 = .0588

Those numbers are only equal after rounding.

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08-19-2020 , 08:33 PM

#12

David Sklansky

Administrator

Join Date: Aug 2002 Posts: 17,096

Quote:

Originally Posted by heehaww

No, these stats aren't useful, partly because your opponents aren't playing blindly or randomly, which means the real probability changes the moment they make a decision.

Not completely happy with this answer. Do you see why?

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08-21-2020 , 03:04 PM

#13

heehaww

Pooh-Bah

Join Date: Aug 2011 Posts: 5,081

More precisely, the stat isn't useful because your opponents' ranges aren't ATC (even if they aren't playing blindly or randomly).

Is that what you're getting at, David?

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08-21-2020 , 06:33 PM

#14

statmanhal

Pooh-Bah

Join Date: Jan 2009 Posts: 4,987

I suppose heehaw is referring to the fact that your opponents can't have the cards you hold. If preflop and you are UTG, I would think assuming ATC is very reasonable

How useful is another issue.

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08-21-2020 , 10:52 PM

#15

heehaww

Pooh-Bah

Join Date: Aug 2011 Posts: 5,081

No I was making a captain obvious point: if a villain raises, that narrows their range and changes any initial probabilities.

Sure when you're UTG or first in, the people behind you have ATC and I suppose if you were building your ranges from scratch with pen and paper, knowing all the initial probabilities would help.

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